Question

Solve the following equation:
$\frac{\log (\sqrt{x+1}+1)}{\log \sqrt[3]{3x}-40}=3$

Answer 1

$\frac{log(\sqrt{x+1}+1}{log\sqrt[3]{3x}-40}=8\Rightarrow log(1+\sqrt{x+1})=logx-120\Rightarrow 1+\sqrt{x+1}=\frac{x}{{10}^{120}}\Rightarrow \sqrt{x+1}=\frac{(x-{10}^{120})}{{10}^{120}}\Rightarrow {10}^{240}(x+1)=x^2-2.{10}^{120}x+{10}^{240}\Rightarrow x^2-(2.{10}^{120}+10^{240})x=0\Rightarrow x=0orx=2.{10}^{120}+{10}^{240}$

But $x\ne 0$

$x={10}^{120}(2+{10}^{120}).$